Answer

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**Hint:**In this problem we have to evaluate $\cos 150{}^\circ$. For this type of problem, we will write the given angle as the sum or difference of two angles for which we have the values of the trigonometric ratios. Now according to the trigonometric ratio, we will use the appropriate formula and simplify the equation to get the result. For this problem, we will write the given angle $150{}^\circ$ as the sum of the $90{}^\circ$, $60{}^\circ$. Now we will use the trigonometric formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$. In this formula, we will apply all the trigonometric values we have and simplify them to get the required result.

**Formula use:**

1. we have the trigonometric formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$.

**Complete step by step solution:**

Given that, $\cos 150{}^\circ$.

Consider the given angle $150{}^\circ$ . We can write the given angle $150{}^\circ$ as sum of the $90{}^\circ$ and $60{}^\circ$ . Mathematically we can write

$150{}^\circ =90{}^\circ +60{}^\circ$

Applying the trigonometric function $\cos$ on both sides of the above equation, then we will get

$\Rightarrow \cos 150{}^\circ =\cos \left( 60{}^\circ +90{}^\circ \right)$

Using the trigonometric formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ in the above equation, then we will get

$\Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=\cos 60{}^\circ .\cos 90{}^\circ -\sin 60{}^\circ \sin 90{}^\circ$

To solve the above equation, we need to have the values of two trigonometric ratios which are $\sin$ and $\cos$.

Considering the $\cos$ terms in the above equation.

We have $\cos 60{}^\circ$ , $\cos 90{}^\circ$.

We know that $\cos 60{}^\circ =\dfrac{1}{2}$ and also, we know that $\cos 90{}^\circ =0$.

Now substituting these values in the above equation, then we will get

$\Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=\dfrac{1}{2}.0-\sin 60{}^\circ \sin 90{}^\circ$

We know that the product of anything with the zero will be zero, then we will get

$\begin{align}

& \Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=0-\sin 60{}^\circ \sin 90{}^\circ \\

& \Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=-\sin 60{}^\circ \sin 90{}^\circ \\

\end{align}$

Considering the $\sin$ terms in the above equation.

We have $\sin 60{}^\circ$ , $\sin 90{}^\circ$.

We know that $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and also, we know that $\sin 90{}^\circ =1$

Substituting these values in the above equation, then we will get

$\Rightarrow \cos \left( 60{}^\circ +90{}^\circ \right)=-\dfrac{\sqrt{3}}{2}\times 1$

**Simplifying the above equation, we have the value of $\cos 150{}^\circ$ as $-\dfrac{\sqrt{3}}{2}$.**

**Note:**

We can find the value of the above expression in another method. We can use property trigonometric table, unit circle and property of complementary arc. That is

$\cos 150{}^\circ =\cos \left( 90{}^\circ +60{}^\circ \right)$

We know that $\cos \left( 90{}^\circ +\theta \right)=-\sin \theta$

$\Rightarrow \cos 150{}^\circ =-\sin 60{}^\circ$

We know that $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ then we will get final result that is

$\Rightarrow \cos 150{}^\circ =-\dfrac{\sqrt{3}}{2}$.

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